#### Answer

$\{\ (1,2),\ (1, -2),\ (-1,2),\ (-1, -2)\ \}$

#### Work Step by Step

1. Eliminating:
eliminate the $(y^{2})$ terms:
$\left\{\begin{array}{lll}
3x^{2}-2y^{2}=-5 & , & \\
2x^{2}-y^{2}=-2 & , & /\times(-2), add
\end{array}\right.$
$(3-4)x^{2}=-5+4$
$-x^{2}=-1$
2. Solving:
$-x^{2}=-1\qquad/\times(-1)$
$x^{2}=1$
$x=\pm 1$
3. Back-substituting into $2x^{2}-y^{2}=-2$:
$\left[\begin{array}{lll}
x=1 & or & x=-1\\
2(1)-y^{2}=-2 & & 2(1)-y^{2}=-2 \\
y^{2}=4 & & y^{2}=4\\
y=\pm 2 & & y=\pm 2
\end{array}\right]$
The solution set is
$\{\ (1,2),\ (1, -2),\ (-1,2),\ (-1, -2)\ \}$